NCERT Solutions for Exercise 3.3 Class 11 Maths Chapter 3 - Trigonometric Functions

NCERT solutions for class 11 maths chapter 3 trigonometric functions-Exercise: 3.3

Question:1 Prove that ${\sin }^2\left(\frac{\pi }{6}\right)+{\cos }^2\left(\frac{\pi }{3}\right)-{\tan }^2\left(\frac{\pi }{4}\right)=-\frac{1}{2}$

Answer:

We know the values of sin (30 degree), cos (60 degree) and tan (45 degree). That is:

$$\begin{gathered} \sin \left(\frac{\pi }{6}\right)=\left(\frac{1}{2}\right) \\ \cos \left(\frac{\pi }{3}\right)=\left(\frac{1}{2}\right) \\ \tan \left(\frac{\pi }{4}\right)=1 \end{gathered}$$
${\sin }^2\frac{\pi }{6}+{\cos }^2\frac{\pi }{3}-{\tan }^2\frac{\pi }{4}=$ ${\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2-1^2$

$=\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}$
= R.H.S.

Question:2 Prove that $2{\sin }^2\left(\frac{\pi }{6}\right)+cose{c}^2\left(\frac{7\pi }{6}\right){\cos }^2\frac{\pi }{3}=\frac{3}{2}$

Answer:

The solutions for the given problem is done as follows.

$$\begin{gathered} \sin \frac{\pi }{6}=\frac{1}{2} \\ cosec\frac{7\pi }{6}=cosec(\pi +\frac{\pi }{6})=-cosec\frac{\pi }{6}=-2 \\ \cos \frac{\pi }{3}=\frac{1}{2} \end{gathered}$$
$$\begin{gathered} 2{\sin }^2\frac{\pi }{6}+cose{c}^2\frac{7\pi }{6}{\cos }^2\frac{\pi }{3}=2{\left(\frac{1}{2}\right)}^2+{(-2)}^2{\left(\frac{1}{2}\right)}^2 \\ \Rightarrow 2\times \frac{1}{4}+4\times \frac{1}{4}=\frac{1}{2}+1=\frac{3}{2} \end{gathered}$$
R.H.S.

Question:3 Prove that ${\cot }^2\left(\frac{\pi }{6}\right)+\csc \left(\frac{5\pi }{6}\right)+3{\tan }^2\left(\frac{\pi }{6}\right)=6$

Answer:

We know the values of cot(30 degree), tan (30 degree) and cosec (30 degree)

$$\begin{gathered} \cot \frac{\pi }{6}=\sqrt{3} \\ cosec\frac{5\pi }{6}=cosec(\pi -\frac{\pi }{6})=cosec\frac{\pi }{6}=2 \\ \tan \frac{\pi }{6}=\frac{1}{\sqrt{3}} \end{gathered}$$

$$\begin{gathered} {\cot }^2\frac{\pi }{6}+cosec\frac{5\pi }{6}+3{\tan }^2\frac{\pi }{6}={(\sqrt{(}3))}^2+2+3\times {\left(\frac{1}{\sqrt{3}}\right)}^2 \\ \Rightarrow 3+2+1=6 \end{gathered}$$
R.H.S.

Question:4 Prove that $2{\sin }^2\left(\frac{3\pi }{4}\right)+2{\cos }^2\left(\frac{\pi }{4}\right)+2{\sec }^2\left(\frac{\pi }{3}\right)=10$

Answer:

$$\begin{gathered} \sin \frac{3\pi }{4}=\sin (\pi -\frac{\pi }{4})=\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}} \\ \cos \frac{\pi }{4}=\frac{1}{\sqrt{2}} \\ \sec \frac{\pi }{3}=2 \end{gathered}$$
Using the above values

$$\begin{gathered} 2{\sin }^2\frac{3\pi }{4}+2{\cos }^2\frac{\pi }{4}+2{\sec }^2\frac{\pi }{3}=2\times {\left(\frac{1}{\sqrt{2}}\right)}^2+2\times {\left(\frac{1}{\sqrt{2}}\right)}^2+2{\left(2\right)}^2 \\ \Rightarrow 1+1+8=10 \end{gathered}$$
R.H.S.

Question:5(i) Find the value of $(i)\sin {75}^{\circ }$

Answer:

$\sin {75}^{\circ }=\sin ({45}^{\circ }+{30}^{\circ })$
We know that
(sin(x+y)=sinxcosy + cosxsiny)
Using this idendity

$$\begin{gathered} \sin {75}^{\circ }=\sin ({45}^{\circ }+{30}^{\circ })=\sin {45}^{\circ }\cos {30}^{\circ }+\cos {45}^{\circ }\sin {30}^{\circ } \\ \Rightarrow \frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2} \\ \Rightarrow \frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}=\frac{\sqrt{3}+1}{2\sqrt{2}} \end{gathered}$$

Question:5(ii) Find the value of
$(ii)\tan {15}^{\circ }$

Answer:

$\tan {15}^{\circ }=\tan ({45}^{\circ }-{30}^{\circ })$
We know that,

$[\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}]$
By using this we can write

$$\begin{gathered} \tan ({45}^{\circ }-{30}^{\circ })=\frac{\tan {45}^{\circ }-tan{30}^{\circ }}{1+\tan {45}^{\circ }\tan {30}^{\circ }} \\ \Rightarrow \frac{1-\frac{1}{\sqrt{3}}}{1+1\left(\frac{1}{\sqrt{3}}\right)}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{{(\sqrt{3}-1)}^2}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{3+1-2\sqrt{3}}{{\left(\sqrt{3}\right)}^2-{\left(1\right)}^2} \\ \Rightarrow \frac{4-2\sqrt{3}}{3-1}=\frac{2(2-\sqrt{3})}{2}=2-\sqrt{3} \end{gathered}$$

Question:6 Prove the following: $\cos (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-y)-\sin (\frac{\pi }{4}-x)\sin (\frac{\pi }{4}-y)=\sin (x+y)$

Answer:

$\cos (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-y)-\sin (\frac{\pi }{4}-x)\sin (\frac{\pi }{4}-y)$

Multiply and divide by 2 both cos and sin functions
We get,

$\frac{1}{2}[2\cos (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-y)]+\frac{1}{2}[-2\sin (\frac{\pi }{4}-x)\sin (\frac{\pi }{4}-y)]$

Now, we know that

2cosAcosB = cos(A+B) + cos(A-B) -(i)
-2sinAsinB = cos(A+B) - cos(A-B) -(ii)
We use these two identities

In our question A = $(\frac{\pi }{4}-x)$

B = $(\frac{\pi }{4}-y)$
So,

$$\begin{gathered} \frac{1}{2}[\cos \{(\frac{\pi }{4}-x)+(\frac{\pi }{4}-y)\}+\cos \{(\frac{\pi }{4}-x)-(\frac{\pi }{4}-y)\}]+ \\ \frac{1}{2}[\cos \{(\frac{\pi }{4}-x)+(\frac{\pi }{4}-y)\}-\cos \{(\frac{\pi }{4}-x)+(\frac{\pi }{4}-y)\}] \end{gathered}$$

$\Rightarrow 2\times \frac{1}{2}[\cos \{(\frac{\pi }{4}-x)+(\frac{\pi }{4}-y)\}]$

$=\cos [\frac{\pi }{2}-(x+y)]$

As we know that

$(\cos (\frac{\pi }{2}-A)=\sin A)$
By using this

$=\cos [\frac{\pi }{2}-(x+y)]$ $=\sin (x+y)$

R.H.S

Question:7 Prove the following $\frac{\tan (\frac{\pi }{4}+x)}{\tan (\frac{\pi }{4}-x)}={\left(\frac{1+\tan x}{1-\tan x}\right)}^2$

Answer:

As we know that

$(\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B})$ and $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

So, by using these identities

$\frac{\tan (\frac{\pi }{4}+x)}{\tan (\frac{\pi }{4}-x)}=\frac{\frac{\tan \frac{\pi }{4}+\tan x}{1-\tan \frac{\pi }{4}\tan x}}{\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\tan x}}=\frac{\frac{1+\tan x}{1-\tan x}}{\frac{1-\tan x}{1+\tan x}}={\left(\frac{1+\tan x}{1-\tan x}\right)}^2$
R.H.S

Question:8 Prove the following $\frac{\cos (\pi +x)\cos (-x)}{\sin (\pi -x)\cos (\frac{\pi }{2}+x)}={\cot }^{2}x$

Answer:

As we know that,
$\cos (\pi +x)=-\cos x$ , $\sin (\pi -x)=\sin x$ , $\cos (\frac{\pi }{2}+x)=-\sin x$
and
$\cos (-x)=\cos x$

By using these our equation simplify to

$\frac{\cos x\times -\cos x}{sinx\times -\sin x}=\frac{-{\cos }^{2}x}{-{\sin }^{2}x}={\cot }^{2}x$ $(\because \cot x=\frac{\cos x}{\sin x})$
R.H.S.

Question:9 Prove the following $\cos (\frac{3\pi }{2}+x)\cos (2\pi +x)[\cot (\frac{3\pi }{2}-x)+\cot (2\pi +x)]=1$

Answer:

We know that

$$\begin{gathered} \cos (\frac{3\pi }{2}+x)=\sin x \\ \cos (2\pi +x)=\cos x \\ \cot (\frac{3\pi }{2}-x)=\tan x \\ \cot (2\pi +x)=\cot x \end{gathered}$$

So, by using these our equation simplifies to

$$\begin{gathered} \cos (\frac{3\pi }{2}+x)\cos (2\pi +x)[\cot (\frac{3\pi }{2}-x)+\cot (2\pi +x)] \\ =\sin x\cos x[\tan x+\cot x]=\sin x\cos x[\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}] \\ \Rightarrow \sin x\cos x\left[\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}\right]={\sin }^{2}x+{\cos }^{2}x=1 \end{gathered}$$ R.H.S.

Question:10 Prove the following $\sin (n+1)x\sin (n+2)x+\cos (n+1)x\cos (n+2)x=\cos x$

Answer:

Multiply and divide by 2

$=\frac{2\sin (n+1)x\sin (n+2)x+2\cos (n+1)x\cos (n+2)x}{2}$

Now by using identities

-2sinAsinB = cos(A+B) - cos(A-B)
2cosAcosB = cos(A+B) + cos(A-B)

$$\begin{gathered} \frac{\{-(\cos (2n+3)x-\cos (-x))+(\cos (2n+3)+\cos (-x))\}}{2} \\ (\because \cos (-x)=\cos x) \\ =\frac{2\cos x}{2}=\cos x \end{gathered}$$

R.H.S.

Question:11 Prove the following $\cos (\frac{3\pi }{4}+x)-\cos (\frac{3\pi }{4}-x)=-\sqrt{2}\sin x$

Answer:

We know that

[ cos(A+B) - cos (A-B) = -2sinAsinB ]

By using this identity

$$\begin{gathered} \cos (\frac{3\pi }{4}+x)-\cos (\frac{3\pi }{4}-x)=-2\sin \frac{3\pi }{4}\sin x=-2\times \frac{1}{\sqrt{2}}\sin x \\ =-\sqrt{2}\sin x \end{gathered}$$ R.H.S.

Question:12 Prove the following ${\sin }^{2}6x-{\sin }^{2}4x=\sin 2x\sin 10x$

Answer:

We know that
$a^2-b^2=(a+b)(a-b)$

So,
${\sin }^{2}6x-{\sin }^{2}4x=(\sin 6x+\sin 4x)(\sin 6x-\sin 4x)$

Now, we know that

$$\begin{gathered} \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right) \\ \sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right) \end{gathered}$$
By using these identities
sin6x + sin4x = 2sin5x cosx
sin6x - sin4x = 2cos5x sinx

$\Rightarrow {\sin }^{2}6x-{\sin }^{2}4x=(2\cos 5x\sin 5x)(2\sin x\cos x)$

Now,

2sinAcosB = sin(A+B) + sin(A-B)
2cosAsinB = sin(A+B) - sin(A-B)

by using these identities

2cos5x sin5x = sin10x - 0
2sinx cosx = sin2x + 0

hence
${\sin }^{2}6x-{\sin }^{2}4x=\sin 2x\sin 10x$

Question:13 Prove the following ${\cos }^{2}2x-{\cos }^{2}6x=\sin 4x\sin 8x$

Answer:

As we know that

$a^2-b^2=(a-b)(a+b)$

$\therefore {\cos }^{2}2x-{\cos }^{2}6x=(\cos 2x-\cos 6x)(\cos 2x+\cos 6x)$
Now
$$\begin{gathered} \cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right) \\ \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right) \end{gathered}$$
By using these identities

cos2x - cos6x = -2sin(4x)sin(-2x) = 2sin4xsin2x ( $\because$ sin(-x) = -sin x
cos(-x) = cosx)
cos2x + cos 6x = 2cos4xcos(-2x) = 2cos4xcos2x

So our equation becomes

R.H.S.

Question:14 Prove the following $\sin 2x+2\sin 4x+\sin 6x=4{\cos }^{2}x\sin 4x$

Answer:

We know that
$\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$
We are using this identity
sin2x + 2sin4x + sin6x = (sin2x + sin6x) + 2sin4x

sin2x + sin6x = 2sin4xcos(-2x) = 2sin4xcos(2x) ( $\because$ cos(-x) = cos x)

So, our equation becomes
sin2x + 2sin4x + sin6x = 2sin4xcos(2x) + 2sin4x
Now, take the 2sin4x common
sin2x + 2sin4x + sin6x = 2sin4x(cos2x +1) ( $\because \cos 2x=2{\cos }^{2}x-1$ )
=2sin4x( $2{\cos }^{2}x-1$ +1 )
=2sin4x( $2{\cos }^{2}x$ )
= $4\sin 4x{\cos }^{2}x$
R.H.S.

Question:15 Prove the following $\cot 4x(\sin 5x+\sin 3x)=\cot x(\sin 5x-\sin 3x)$

Answer:

We know that
$\sin x+\sin y=2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)$
By using this , we get

sin5x + sin3x = 2sin4xcosx

$\frac{\cos 4x}{\sin 4x}(2\sin 4x\cos x)=2\cos 4x\cos x$

now nultiply and divide by sin x

$$\begin{gathered} \frac{2\cos 4x\cos x\sin x}{\sin x} \\ =\cot x(2\cos 4x\sin x)(\because \frac{\cos x}{sinx}=\cot x) \end{gathered}$$

Now we know that

$2\cos x\sin y=\sin (x+y)-\sin (x-y)$

By using this our equation becomes

$=\cot x(\sin 5x-sin3x)$
R.H.S.

Question:16 Prove the following $\frac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=-\frac{\sin 2x}{\cos 10x}$

Answer:

As we know that

$$\begin{gathered} \cos x-\cos y=-2\sin \frac{x+y}{2}\sin \frac{x-y}{2} \\ \cos 9x-\cos 5x=-2\sin 7x\sin 2x \\ \sin x-\sin y=2\cos \frac{x+y}{2}\sin \frac{x-y}{2} \\ \sin 17x-\sin 3x=2\cos 10x\sin 7x \\ \frac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=\frac{-2\sin 7x\sin 2x}{2\cos 10x\sin 7x}=-\frac{\sin 2x}{\cos 10x} \end{gathered}$$
R.H.S.

Question:17 Prove the following $\frac{\sin 5x+\sin 3x}{\cos 5x+\cos 3x}=\tan 4x$

Answer:

We know that

$$\begin{gathered} \sin A+\sin B=2\sin \frac{A+B}{2}\cos \frac{A-B}{2} \\ and \\ \cos A+\cos B=2\cos \frac{A+B}{2}\cos \frac{A-B}{2} \end{gathered}$$

We use these identities

$$\begin{gathered} \sin 5x+\sin 3x=2\sin 4x\cos x \\ \cos 5x+\cos 3x=2\cos 4x\cos x \\ \frac{\sin 5x+\sin 3x}{\cos 5x+\cos 3x}=\frac{2\sin 4x\cos x}{2\cos 4x\cos x}=\frac{\sin 4x}{\cos 4x}=\tan 4x \end{gathered}$$
R.H.S.

Question:18 Prove the following $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{(x-y)}{2}$

Answer:

We know that
$$\begin{gathered} \sin x-\sin y=2\cos \frac{x+y}{2}\sin \frac{x-y}{2} \\ and \\ \cos x+\cos y=2\cos \frac{x+y}{2}\cos \frac{x-y}{2} \end{gathered}$$

We use these identities

$$\begin{gathered} Weusetheseidentities \\ \frac{\sin x-\sin y}{\cos x+\cos y}=\frac{2\cos \frac{x+y}{2}\sin \frac{x-y}{2}}{2\cos \frac{x+y}{2}\cos \frac{x-y}{2}}=\frac{\sin \frac{x-y}{2}}{\cos \frac{x-y}{2}}=\tan \frac{x-y}{2} \end{gathered}$$

R.H.S.

Question:19 Prove the following $\frac{\sin x+\sin 3x}{\cos x+\cos 3x}=\tan 2x$

Answer:

We know that

$$\begin{gathered} \sin x+\sin y=2\sin \frac{x+y}{2}\cos \frac{x-y}{2} \\ and \\ \cos x+\cos y=2\cos \frac{x+y}{2}\cos \frac{x-y}{2} \\ Weusetheseequations \\ \sin x+\sin 3x=2\sin 2x\cos (-x)=2\sin 2x\cos x(\because \cos (-x)=\cos x) \\ \cos x+\cos 3x=2\cos 2x\cos (-x)=2\cos 2x\cos x(\because \cos (-x)=\cos x) \\ \frac{\sin x+\sin 3x}{\cos x+\cos 3x}=\frac{2\sin 2x\cos x}{2\cos 2x\cos x}=\frac{\sin 2x}{\cos 2x}=\tan 2x \end{gathered}$$ R.H.S.

Question:20 Prove the following $\frac{\sin x-\sin 3x}{{\sin }^{2}x-{\cos }^{2}x}=2\sin x$

Answer:
We know that

$$\begin{gathered} \sin 3x=3\sin x-4{\sin }^{3}x,{\cos }^2-{\sin }^{2}x=\cos 2x \\ and \\ \cos 2x=1-2{\sin }^{2}x \end{gathered}$$

We use these identities

$$\begin{gathered} \sin x-\sin 3x=\sin x-(3\sin x-4{\sin }^{3}x)=4{\sin }^{3}x-2\sin x \\ .=2\sin x(2{\sin }^{2}x-1) \\ {\cos }^{2}x-{\sin }^2=\cos 2x \\ \cos 2x=1-2{\sin }^{2}x \end{gathered}$$ $$\begin{gathered} \sin x-\sin 3x=\sin x-(3\sin x-4{\sin }^{3}x)=4{\sin }^{3}x-2\sin x \\ .=2\sin x(2{\sin }^{2}x-1) \\ {\sin }^2-{\cos }^{2}x=-\cos 2x(\cos 2x=1-2{\sin }^{2}x) \\ {\sin }^2-{\cos }^{2}x=-(1-2{\sin }^{2}x)=2{\sin }^{(}2)x-1 \\ \frac{\sin x-\sin 3x}{{\sin }^2-{\cos }^{2}x}=\frac{2\sin x(2{\sin }^{2}x-1)}{2{\sin }^{(}2)x-1}=2\sin x \end{gathered}$$ $$\begin{gathered} \sin x-\sin 3x=\sin x-(3\sin x-4{\sin }^{3}x)=4{\sin }^{3}x-2\sin x \\ .=2\sin x(2{\sin }^{2}x-1) \\ {\sin }^2-{\cos }^{2}x=-\cos 2x(\because \cos 2x=1-2{\sin }^{2}x) \\ {\sin }^2-{\cos }^{2}x=-(1-2{\sin }^{2}x)=2{\sin }^{(}2)x-1 \\ \frac{\sin x-\sin 3x}{{\sin }^2-{\cos }^{2}x}=\frac{2\sin x(2{\sin }^{2}x-1)}{2{\sin }^{2}x-1}=2\sin x \end{gathered}$$
R.H.S.

Question:21 Prove the following $\frac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}=\cot 3x$

Answer:

We know that

$$\begin{gathered} \cos x+\cos y=2\cos \frac{x+y}{2}\cos \frac{x-y}{2} \\ and \\ \sin x+\sin y=2\sin \frac{x+y}{2}\cos \frac{x-y}{2} \end{gathered}$$
We use these identities

$$\begin{gathered} \frac{(\cos 4x+\cos 2x)+\cos 3x}{(\sin 4x+\sin 2x)+\sin 3x}=\frac{2\cos 3x\cos x+\cos 3x}{2\sin 3x\cos x+\sin 3x}=\frac{2\cos 3x(1+\cos x)}{2\sin 3x(1+\cos x)} \\ =cot3x \end{gathered}$$

=RHS

Question:22 prove the following $\cot x\cot 2x-\cot 2x\cot 3x-\cot 3x\cot x=1$

Answer:

cot x cot2x - cot3x(cot2x - cotx)
Now we can write cot3x = cot(2x + x)

and we know that

$cot(a+b)=\frac{\cot a\cot b-1}{\cot a+\cot b}$
So,
$cotxcot2x-\frac{\cot 2x\cot x-1}{\cot 2x+\cot x}(cot2x+cotx)$
= cotx cot2x - (cot2xcotx -1)
= cotx cot2x - cot2xcotx +1
= 1 = R.H.S.

Question:23 Prove that $\tan 4x=\frac{4\tan x(1-{\tan }^{2}x)}{1-6{\tan }^{2}x+{\tan }^{4}x}$

Answer:

We know that

$tan2A=\frac{2\tan A}{1-{\tan }^{2}A}$

and we can write tan 4x = tan 2(2x)
So, $tan4x=\frac{2\tan 2x}{1-{\tan }^{2}2x}$ = $\frac{2(\frac{2\tan x}{1-{\tan }^{2}x})}{1-(\frac{2\tan x}{1-{\tan }^{2}x}{)}^2}$


= $\frac{2(2\tan x)(1-{\tan }^{2}x)}{(1-\tan x{)}^2-(4{\tan }^{2}x)}$

= $\frac{(4\tan x)(1-{\tan }^{2}x)}{(1{)}^2+({\tan }^{2}x{)}^2-2{\tan }^{2}x-(4{\tan }^{2}x)}$

= $\frac{(4\tan x)(1-{\tan }^{2}x)}{1^2+{\tan }^{4}x-6{\tan }^{2}x}$ = R.H.S.

Question:24 Prove the following $\cos 4x=1-8{\sin }^{2}x{\cos }^{2}x$

Answer:

We know that
$cos2x=1-2{\sin }^{2}x$
We use this in our problem
cos 4x = cos 2(2x)
= $1-2{\sin }^{2}2x$
= $1-2(2\sin x\cos x{)}^2$ $(\because \sin 2x=2\sin x\cos x)$
= $1-8{\sin }^{2}x{\cos }^{2}x$ = R.H.S.

Question:25 Prove the following $\cos 6x=32{\cos }^{6}x-48{\cos }^{4}x+18{\cos }^{2}x-1$

Answer:

We know that
cos 3x = 4 ${\cos }^{3}x$ - 3cos x
we use this in our problem
we can write cos 6x as cos 3(2x)
cos 3(2x) = 4 ${\cos }^{3}2x$ - 3 cos 2x
= $4(2{\cos }^{2}x-1{)}^3$ - $3(2{\cos }^{2}x-1)$ $(\because \cos 2x=2{\cos }^{2}x-1)$
= $4[(2co{s}^{2}x{)}^3-(1{)}^3-3(2co{s}^{2}x{)}^2(1)+3(2co{s}^{2}x)(1{)}^2]$ $-6{\cos }^{2}x+3$ $(\because (a-b{)}^3=a^3-b^3-3a^{2}b+3a{b}^2)$
= 32 $co{s}^{6}x$ - 4 - 48 $co{s}^{4}x$ + 24 $co{s}^{2}x$ - $6{\cos }^{2}x+3$
= 32 $co{s}^{6}x$ - 48 $co{s}^{4}x$ + 18 $co{s}^{2}x$ - 1 = R.H.S.